Q48 / 102 · NTA Abhyas (JEE Main) 2020 · Chemistry · Ionic Equilibrium

AgBr(s)+2S2O32-(aq)⇌Ag(S2O3)23-(aq)+Br-(aq)
Given Ksp(AgBr)=5×10-13;KfAg(S2O3)23-=5×1013
What is the molar solubility of AgBr in 0.1 M Na2S2O3 ?

  • A
    0.5 M
  • B
    0.45 M
  • Correct answer
    C
    0.045 M
  • D
    None of these

Solution

AgBr(s)⇌Ag+(aq)+Br-(aq)Ksp
Ag+(aq)+2S2O32-(aq)⇌Ag(S2O3)23-(aq)Kf
AgBrs+2S2O32-(aq)0.1-2x⇌Ag(S2O3)23-(aq)x+Br-xaq
Here, K=Ksp×Kf=25
K=x2(0.1-2x)2
25=x2(0.1-2x)2
On solving x=0.045 M.

Hence, the correct answer is option: 'C'.

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